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The table contains observed (O) values (data).
Number of hours worked per week by volunteer type (observed)
Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours Row Total
Community College Students 111 96 48 255
Four-Year College Students 96 133 61 290
Nonstudents 91 150 53 294
Column Total 298 379 162 839

Are the number of hours volunteered independent of the type of volunteer?

The observed table and the question at the end of the problem, "Are the number of hours volunteered independent of the type of volunteer?" tell you this is a test ofindependence. The two factors are number of hours volunteered and type of volunteer . This test is always right-tailed.

H o : The number of hours volunteered is independent of the type of volunteer.

H a : The number of hours volunteered is dependent on the type of volunteer.

The expected table is:

The table contains expected ( E ) values (data).
Number of hours worked per week by volunteer type (expected)
Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours
Community College Students 90.57 115.19 49.24
Four-Year College Students 103.00 131.00 56.00
Nonstudents 104.42 132.81 56.77

For example, the calculation for the expected frequency for the top left cell is

E = (row total)(column total) total number surveyed = 255 298 839 = 90.57

Calculate the test statistic: χ 2 = 12.99 (calculator or computer)

Distribution for the test: χ 4 2

df = ( 3 columns - 1 ) ( 3 rows - 1 ) = ( 2 ) ( 2 ) = 4

Graph:

Nonsymmetrical chi-square curve with values of 0 and 12.99 on the x-axis representing the test statistic of number of hours worked by volunteers of different types. A vertical upward line extends from 12.99 to the curve and the area to the right of this is equal to the p-value.

Probability statement: p-value = P ( χ 2 > 12.99 ) = 0.0113

Compare α and the p-value : Since no α is given, assume α = 0.05 . p-value = 0.0113 . α > p-value .

Make a decision: Since α > p-value , reject H o . This means that the factors are not independent.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer aredependent on one another.

For the above example, if there had been another type of volunteer, teenagers, whatwould the degrees of freedom be?

Calculator instructions follow.

TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 3 ENTER 3 ENTER . Enter the table values by row from Example 11-6. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 12.9909 and the p-value = 0.0113 . Do the procedure a second time but arrow down to Draw instead of calculate .

De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measuredanxiety level and need to succeed in school. The table shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independentevents.

Need to succeed in school vs. anxiety level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400

How many high anxiety level students are expected to have a high need to succeed in school?

The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.

E = (row total)(column total) total surveyed = 155 57 400 = 22.09

The expected number of students who have a high anxiety level and a high need to succeed in school is about 22.

If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?

The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.

  • E = (row total)(column total) total surveyed =
  • The expected number of students who have a med-low anxiety level and a low need to succeed in school is about:
  • E = (row total)(column total) total surveyed = 8.19
  • 8
Practice Key Terms 1

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Source:  OpenStax, Quantitative information analysis iii. OpenStax CNX. Dec 25, 2009 Download for free at http://cnx.org/content/col11155/1.1
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